Find electric field a distance from a charged sheet

AI Thread Summary
To find the electric field 9 mm from a charged sheet with a charge density of 5 μC/mm², the formula E=σ/2ε0 is used. The initial calculation yields E=2.8 x 10^5 N/C, but the expected answer is approximately 5.6 x 10^11 N/C. The discrepancy arises from incorrect unit conversions; charge density must be converted to SI units (C/m²) for accurate results. The correct conversion leads to E=2.82 x 10^11 V/m. Understanding the significance of a "sheet of infinitesimal area" is also a point of confusion in the discussion.
rocapp
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Homework Statement



For a sheet of infinitesimal area, find the electric field 9 mm from the surface if the charge density is 5 μC/mm^2.

Homework Equations



E=σ/2ε0

The Attempt at a Solution



E=(5x10^-6 C/m^2)/(2*8.85x10^-12)

= 2.8 x10^5N/C

I know the correct answer is about 5.6x10^11 N/C, but I'm not sure why. Probably units? I'm not sure how to get N/C at the end when I'm given microcoulombs/mm^2 to begin with. Please help! Thank in advance.
 
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Yes, this is a problem with the units.
You need to convert σ into SI units, ie C/m2
 
(5C/m^2) / (2*8.85x10^-12C/Vm) = 2.82x10^11 V/m
 
rocapp said:

Homework Statement



For a sheet of infinitesimal area, find the electric field 9 mm from the surface if the charge density is 5 μC/mm^2.

Homework Equations



E=σ/2ε0

The Attempt at a Solution



E=(5x10^-6 C/m^2)/(2*8.85x10^-12)

= 2.8 x10^5N/C

I know the correct answer is about 5.6x10^11 N/C, but I'm not sure why. Probably units? I'm not sure how to get N/C at the end when I'm given microcoulombs/mm^2 to begin with. Please help! Thank in advance.
What on Earth is a sheet of infinitesimal area ?
 

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